Question 176922
{{{5k^2=3k+2}}} Start with the given equation.



{{{5k^2-3k-2=0}}} Get all terms to the left side.



Notice we have a quadratic equation in the form of {{{ak^2+bk+c}}} where {{{a=5}}}, {{{b=-3}}}, and {{{c=-2}}}



Let's use the quadratic formula to solve for k



{{{k = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{k = (-(-3) +- sqrt( (-3)^2-4(5)(-2) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=-3}}}, and {{{c=-2}}}



{{{k = (3 +- sqrt( (-3)^2-4(5)(-2) ))/(2(5))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{k = (3 +- sqrt( 9-4(5)(-2) ))/(2(5))}}} Square {{{-3}}} to get {{{9}}}. 



{{{k = (3 +- sqrt( 9--40 ))/(2(5))}}} Multiply {{{4(5)(-2)}}} to get {{{-40}}}



{{{k = (3 +- sqrt( 9+40 ))/(2(5))}}} Rewrite {{{sqrt(9--40)}}} as {{{sqrt(9+40)}}}



{{{k = (3 +- sqrt( 49 ))/(2(5))}}} Add {{{9}}} to {{{40}}} to get {{{49}}}



{{{k = (3 +- sqrt( 49 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{k = (3 +- 7)/(10)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{k = (3 + 7)/(10)}}} or {{{k = (3 - 7)/(10)}}} Break up the expression. 



{{{k = (10)/(10)}}} or {{{k =  (-4)/(10)}}} Combine like terms. 



{{{k = 1}}} or {{{k = -2/5}}} Simplify. 



So the answers are {{{k = 1}}} or {{{k = -2/5}}}